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### JEE Main 2017 (Online) 9th April Morning Slot

A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2 . The car will catch up with the bus after a time of :
A
$\sqrt {110} \,s$
B
$\sqrt {120} \,s$
C
$10\,\,\sqrt 2 \,s$
D
15 s

## Explanation

Acceleration of Car, aC = 4 m/s2

Acceleration of bus, aB = 2 m/s2

Initial distance between them, S = 200 m

Acceleration of Car with respect to bus,

aCB = aC $-$ aB = 4 $-$ 2 = 2 m/s2

As initially both are at rest so, uCB = 0

$\therefore\,\,\,$ S = UCB $\times$ t + ${1 \over 2}$ aCB $\times$ t2

$\Rightarrow $$\,\,\, 200 = 0 + {1 \over 2} \times 2 \times t2 \Rightarrow$$\,\,\,$ t2 = 200

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### JEE Main 2018 (Online) 15th April Morning Slot

The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in $15$ $s$ and (ii) the time at which the car will catch up with the scooter are, respectively.

A
$112.5$ $m$ and $22.5$ $s$
B
$337.5$ $m$ and $25$ $s$
C
$112.5$ $m$ and $15$ $s$
D
$225.5$ $m$ and $10$ $s$

## Explanation

Till 15 sec cash has accelerative motion and scooter has constant velocity in entire motion.

Total Distance travelled by the car in 15 sec, = ${1 \over 2}$ $\times$ ${{45} \over {15}}$ $\times$ (15)2 = ${{675} \over 2}$ m

Distance travelled by scooter in 15 sec.

= V $\times$ t = 30 $\times$ 15 = 450 m

$\therefore\,\,\,\,$ Difference between distance travelled by the car and scooter in 15 sec,

= 450 $-$ ${{675} \over 2}$ = 112.5 m

Now, assume car catches the scooter in time t,

$\therefore\,\,\,\,$ Car travelled in t sec = scooter travelled in t sec.

$\Rightarrow$ $\,\,\,\,$ ${{675} \over 2}$ + 45(t$-$ 15) = 30 $\times$ t

$\Rightarrow$ $\,\,\,\,$ 337.5 + 45t $-$ 675 = 30t

$\Rightarrow$ $\,\,\,\,$ 15 t = 337.5

$\Rightarrow$ $\,\,\,\,$ t = 22.5 sec.